Final Answer:
The expected value \(E[X]\) for a random variable \(X\) with a probability density function \(f\) that is an even function is zero. It is not possible for \(f\) to be a probability density function and odd simultaneously, as an odd function would violate the non-negativity property required for a valid probability density function.
Step-by-step explanation:
To compute \(E[X]\) for a random variable with an even probability density function, we use the formula:
\[ E[X] = \int_{-\infty}^{\infty} x f(x) \, dx \]
Since \(f\) is an even function (\(f(-x) = f(x)\)), the integral simplifies to:
\[ E[X] = 2 \int_{0}^{\infty} x f(x) \, dx \]
However, due to the even symmetry, the integral of an odd function over a symmetric interval is zero, resulting in \(E[X] = 0\).
Regarding the second part, for a function to be a probability density function, it must satisfy two key conditions: it should be non-negative for all values and the total area under the curve should be equal to 1. An odd function, by definition, does not satisfy the non-negativity condition for all \(x\) values, as it changes sign across the origin.
Therefore, it cannot represent a valid probability density function. The fundamental properties of probability density functions require them to be non-negative, and this property is violated by odd functions, making it impossible for an odd function to serve as a probability density function.